2017 2nd International Conference on Artificial Intelligence and Engineering Applications (AIEA 2017)
ISBN: 978-1-60595-485-1
Discussion on Calculation of Cooling Coefficient in
Thermal Teaching Materials
HAIYAN YANG, XINGJU DANG, XIAOPAN LI, HAITAO YANG and YAN CAI
ABSTRACT
The cooling coefficient ε of the reverse cycle is equal to the net heat Q absorbed by the system from the low temperature heat source by one cycle. The net power A of the system is calculated by subtracting the heat Q released by the system to the low temperature heat source in one cycle. The cooling coefficient ε can be greater than 1. It can also be less than 0. The cooling coefficient of the reverse cycle is less than the cooling coefficient of the Carnot cycle.
KEYWORDS
Reverse cycle, Cooling coefficient, Carnot cycle, Inverse Stirling cycle, Joule circulation.
INTRODUCTION
In the domestic higher education thermo logy teaching materials such as [1-4], the calculation of the reversible cycle cooling coefficient cannot be justified. For example, it is assumed that the cooling coefficient of the inverse Sterling cycle equals to the cooling factor of the Carnot reverse cycle. This has brought confusion to teaching.
The more reasonable explanation for the cooling coefficient of the inverse Sterling cycle is equal to the cooling factor of the Carnot reverse cycle is that the process of the endothermic and the exothermic heat of the regenerator is carried out in the system, so that the influence of the outside world. However, according to this interpretation, it should be the actual sub-cycle process approximation as two adiabatic process rather than two equal process, so that the cooling coefficient of natural and Kano reverse cycle of the same. We believe that the cooling coefficient of the inverse Sterling cycle should be less than the cooling factor of the Carnot reverse cycle, since the heat efficiency of the Sterling cycle is lower than that of the Carnot cycle.
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Haiyan Yang, [email protected], Xingju Dang, [email protected], Xiaopan Li, [email protected], Haitao Yang, [email protected], College of Physics and Information Engineering, Zhaotong University, Zhaotong, Yunnan. China.
To this end on the calculation of the cooling coefficient of the following views, I would like to teach at home.
CALCULATION OF EFFICIENCY OF CIRCULATING HEAT ENGINE
The second law of thermodynamics determines a system for reversible positive circulation of the sub-process, there must be a sub-process I is the system to the outside world work, another sub-process II is the outside world to do work on the system; a sub-process III is the system to absorb heat. Another sub-process IV is the system to the outside world to release heat. When the formula η = A / Q is used to calculate the efficiency of the heat engine, the A in the formula is the net work done by the system to the outside in the whole cycle, that is, the process I system subtracts from the outside world. Only the former and not the latter. The Q in the formula is only the heat absorbed by process III, because if it is calculated by the exothermic of process IV, the endotherm of the whole cycle is equal to the work done by the first law of thermodynamics, and the heat efficiency is lost significance.
Analysis of Calculation of Reverse Circulation Cooling Coefficient
Refrigeration Coefficient of Inverse Sterling Cycle.
Inverse Sterling Cycle consists of two isothermal processes and two adjacency processes. In the process of isothermal expansion, the system absorbs the heat of the cold trap to do work on the outside world. Thus, the temperature of the system in this process is slightly lower than the temperature of the cold trap. Because of one, according to the second law of thermodynamics, it is impossible in the outside world does not work on the system under the conditions of heat from high temperature objects to low temperature objects; Second, the heat is due to the existence of temperature difference and the transmission of energy. So isothermal "and so on", but the physical approximation is equal, rather than the mathematical rigor of equality. In the isothermal process, the system temperature is only approximately equal to the thermal pool temperature, rather than strictly equal to the thermal library temperature.
Followed by the body temperature, the system on the outside work done 0, the internal energy becomes larger, it must be to the outside heat. Since the system temperature is higher than the cold trap temperature, it is impossible to continue to heat the heat trap, the heat absorbed must come from the thermal barrier. Due to endotherm, the system temperature continues to rise until it is slightly higher than the temperature of the thermal barrier.
And then isothermal compression, the outside world on the system work, the system heat to the heat barrier. In the last sub-process and other body cooling, the system on the outside work done 0, the system can be reduced, so the heat to the outside world. As the system temperature is lower than the thermal barrier and higher than the cold trap, so the system cannot be released heat absorption and must be absorbed by the cold trap. Due to exothermic, the system temperature continues to decrease until it is slightly below the cold trap temperature. This completes a cycle.
Through the above analysis, we can see that in an Inverse Sterling Cycle, the net heat Q absorbed by the system from the cold trap should be the difference between the heat absorbed during isothermal expansion and the heat released during the cooling process. Therefore, for the work of the material as a single atomic ideal gas, compression before and after the volume of V1 and V2, respectively, the thermal barrier and cold trap thermodynamic temperature were Th and Tc Inverse Sterling Cycle, assuming that the system done by the outside world is fully recycled, then
1
c h c
2
3
ln ( )
2 V
Q R T R T T
V
(1)
1
h c
2
( ) ln V
A R T T
V
(2)
c
h c 1 2
3 2 ln ( ) T
Q
A T T V V
(3)
1 23 Th Tc 2 lnTc V V
(4)
(For example, Th>1.5Tc, but V2>0.5V1), its cooling coefficient becomes negative, the cycle will lead to cold trap becomes more and more hot.
The cooling coefficients of the reverse cycle consisting of two isothermal processes and two isobaric processes.
Several thermal textbooks give the same question that the refrigeration coefficient of the inverse cycle consisting of two isothermal processes and two isobaric processes is equal to the cooling factor of the Kanuho reverse cycle. We think that this exercise is flawed, the two reverse cycle of the cooling coefficient is not equal.
Similar to the Inverse Sterling Cycle, the system absorbs heat from the cold trap during isothermal expansion and exothermic to the thermal barrier during isothermal compression. In the process of isobaric expansion, the system temperature is lower than the temperature of the cold trap and is lower than the temperature of the thermal barrier. Therefore, it can only absorb heat from the heat barrier. In the process of isotactic compression, the system temperature is lower than the temperature Well temperature, so can only be exothermic to cold trap. When the working material is a single atom ideal gas, the pressure before and after isothermal compression is p1 and p2 respectively, and the thermodynamic temperature of the heat barrier and the cold trap are Th and Tc respectively. Assuming that the system is fully utilized in the external work,
2 5
ln 1 2
c h c
p
Q RT R T T
p
(5)
21
ln h c
p
A R T T
p
(6)
2 1
5
2 ln c
h c T Q
p A T T
p
(7)
The cooling coefficient is also lower than the Kano reverse cycle operating between the same two heat sources. And when
21
5 h c 2 lnc p
T T T
p
(8)
(For example, Th>1.3Tc, but p1>0.5p2), its cooling coefficient becomes negative, the cycle will lead to cold trap becomes more and more hot.
The Cooling Factor of the Joule Cycle.
isobaric processes. The system absorbs heat from the cold trap during isobaric expansion and exothermic to the thermal barrier during the isobaric compression process. The external work done on the system is equal to the difference between the heat released by the system and the heat absorbed from the cold trap.
Assuming that the minimum temperature of the system is T1 in a cycle, the corresponding volume is V1, the pressure is p1 during the isobaric expansion, the volume is V2 at the end of the isobaric expansion, and the pressure is p2 during the isobaric compression process. For the single atom ideal gas, the isentropic index γ = 5 / 3. The system on the outside functions are recycled.
The temperature at the end of the isobaric expansion is:
1 1 2 2 T V V
T (10)
The temperature at the end of adiabatic compression is:
1 1/ 0.4
2 2 2
3 2 1
1 1 1
p V p
T T T
p V p
(11)
The temperature at which the adiabatic expansion begins is:
1 1/ 0.4
2 2
4 1 1
1 1
p p
T T T
p p
(12)
2
p 2 1 1
1 5 ( 1) 2
V
Q C T T R T
V
(13)
p 3 4 2 1
0.4 0.4
2 2 2 2
1
1 1 1 1
0.4 2 2 1 1 1 5 1 2 5 1 1 2
A C T T T T
p V p V
R T
p V p V
p V R T p V (14) 1 0.4
2 1 2
1 4 1 3 2
1
p T T
Q
A p T T T T
(15)
The cooling coefficient is equal to the cooling coefficient of the Carnot cycle, which is operated at a cold trap temperature of (T1+T2)/2 and a thermal barrier temperature of (T3+T4)/2. This is due to the existence of two reversible adiabatic processes in the process.
CONCLUSION
heat source to the high heat source heat transfer. In the reverse cycle, this net heat is usually positive, but may also be zero or even become negative, then the reverse cycle cannot achieve the purpose of cooling or pump heat. And when the system cannot effectively use the outside world, the cycle of the cooling coefficient will be smaller.
REFERENCES
1. K.H. Zhao, W.Y. Luo: New-Concept Physics Textbook Thermology (Higher Education Press, China 2005), p. 157 (In Chinese).
2. S.R. Liang, C.N. Liu, Z.H. Sheng: The 2nd book of General Physics Thermology (Higher Education Press, China 2006), p. 105 (In Chinese).
3. C.Li, L.Y. Zhang, S.W. Qian: Thermology (Higher Education Press, China, 2008), p. 135 (In Chinese).